I am brand new to ODE's, and have been having difficulties with this practice problem. Find a 1-parameter solution to the homogenous ODE:$$2xy \, dx+(x^2+y^2) \, dy = 0$$assuming the coefficient of $dy \ne 0$
The textbook would like me to use the subsitution $x = yu$ and $dx=y \, du + u \, dy,\ y \ne 0$
Rewriting the equation with the subsititution:
$$2uy^2(y \, du + u \, dy)+(x^2+y^2) \, dy = 0$$
divide by $y^2$
$$2u(y \, du + u \, dy) + (u^2+1 ) \, dy=0$$
but after further simplification I end up getting ${dy \over y}$ which would mean I would get a logarithm after integrating, and the answer is given as $$3x^2y+y^3 = c$$
Could I get some help/hints as to how this answer was obtained?

The integral of $dy/y$ is $ln(|y|)$ and the integral of $2u/(3u^2+ 1)du$, using the substitution $v= 3u^2+ 1$ so that $dv= 6udu$ or $2udu= dv/3$, is the integral of $1/v dv/3$ which is $$(1/3)ln(|v|)= (1/3)ln(3u^2+ 1)= (1/3)ln(3x^2/y^2+ 1)$$